Question

A ball is thrown upwards with a speed of 39.2 m/s. Calculate (a) the maximum height it reaches, and (b) the time taken in reaching the maximum height.

Answer 1

intial velocity(u) = 39.2 m/s
final velocity(v) = 0

acceleration(a) = -9.8

a) using third equation of motion

2as = v^2 -u^2
2 (-9.8)(s)= (0)^2 - (39.2)^2
-19.6 s = - 1536.64
19.6s = 1536.64
s = 1536.64 / 19.6
= 78.4


so,
the maximum height covered = 78.4 m



b) let time taken be t

using acceleration formula

a= (v-u)/t
-9.8 = (0- 39.2)/t
-9.8 t = -39.2
9.8t = 39.2
t = 39.2 / 9.8
= 4



so,
the time taken was 4s to reach maximum height.

Answer 2

Here is your answer

Hope it will help!